# rate of change according to time problem

• March 3rd 2013, 07:44 AM
mido22
rate of change according to time problem
A prism of an equilateral triangle base , its height is triple the length of its base if the length of its base increases by rate 0.01 cm/sec . Find the rate of increase of volume of the prism at length of base 5 root3 cm
• March 3rd 2013, 10:36 AM
HallsofIvy
Re: rate of change according to time problem
I'm not sure what you mean by "the length of its base". The obvious thing is the length of a side of the triangle but another possibility is the length of an altitude.

First, can you write the formula for volume of such a prism with base side length s and height h?
• March 3rd 2013, 12:06 PM
mido22
Re: rate of change according to time problem
Quote:

Originally Posted by HallsofIvy
I'm not sure what you mean by "the length of its base". The obvious thing is the length of a side of the triangle but another possibility is the length of an altitude.

First, can you write the formula for volume of such a prism with base side length s and height h?

the base length means -->the length of a side of the triangle

here is my attempt :

volume of prism = area of base * height .....assume that base length = x ,, height of prism = h = 3x
so
$v= (\frac{1}{2} \times X \times \frac{\sqrt{3}}{2}X )\times 3X$
$v = \frac{3 \sqrt{3}}{4} \times x^3$by taking derivative with respect to time
so
$\;\frac {dv}{dt} = \frac{9 \sqrt{3}}{4} \times x^2 \times \frac{dx}{dt}$ by substituting i found that
$\frac {dv}{dt} = \frac{27 \sqrt{3}}{16}$

is that right?????(Worried)